At a distance of 30 m the noise from the engine of an jet has an intensity of 130 dB.
At this level, you will be in pain and your ears will hurt. That's why this intensity is know as the "pain threshold".
How far do you have to be from a jet (total distance), in order for the noise to drop down in intensity to 62.2 dB, a level comparable to that of a spoken conversation?
You are after a 67.8 dB reduction, which is a reduction by a factor of 6*10^6 in sound power per area (intensity). You will have to be 2400 times farther away.
Im doing this question right now and I cant understand it! how did you get that value?!
To determine the distance required for the noise intensity to drop to a specific level, we can use the inverse square law formula:
I1/I2 = (d2^2 / d1^2)
Where:
I1 is the initial intensity (130 dB),
I2 is the final intensity (62.2 dB),
d1 is the initial distance (30 m),
d2 is the final distance (which we need to calculate).
Let's plug in the values and solve for d2:
62.2 dB / 130 dB = (d2^2 / 30^2)
0.4782 = (d2^2 / 900)
Cross multiply and solve for d2:
d2^2 = 900 * 0.4782
d2^2 = 430.38
d2 ≈ √430.38
d2 ≈ 20.75 m
Therefore, to reduce the noise intensity from 130 dB to 62.2 dB, you would need to be approximately 20.75 meters away from the jet.